Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 1

Answer

The tension in the third rope is 94.3 N and it points at an angle of $58.0^{\circ}$ below the horizontal.

Work Step by Step

The vertical component of $T_3$ must be equal in magnitude to $T_2$. The horizontal component of $T_3$ must be equal in magnitude to $T_1$. We can find the magnitude of $T_3$ as; $T_3 = \sqrt{(T_{3y})^2+(T_{3x})^2}$ $T_3 = \sqrt{(80~N)^2+(50~N)^2}$ $T_3 = 94.3~N$ We then find the angle $\theta$ below the horizontal; $tan(\theta) = \frac{T_{3y}}{T_{3x}}$ $\theta = arctan(\frac{80~N}{50~N})$ $\theta = 58.0^{\circ}$ The tension in the third rope is 94.3 N and it points at an angle of $58.0^{\circ}$ below the horizontal.
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